Answer:
sun
Explanation:
because all plants use energy from the sun to make food and grow
Answer:
1.8 moles of O₂
Explanation:
The balance chemical equation for said double replacement (photosynthesis) reaction is as follow;
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
According to balance chemical equation,
6 moles of O₂ are produced by = 6 moles of CO₂
So,
1.8 moles of O₂ will be produced by = X moles of O₂
Solving for X,
X = 1.8 mol × 6 mol / 6 mol
X = 1.8 moles of O₂
Stoichiometric problems in which moles are given and moles or other reactant or product asked are the simplest problems. One should only write the balanced chemical equation and perform above method to find the required moles.
The first step is to find the number of moles of OH⁻ that reacted with the HCl. To do this multiply 2.00L by 1.50M to get 3 moles of Ca(OH)₂. Then you multiply 3 by 2 (there are 2 moles of OH⁻ per every 1 mole of Ca(OH)₂) to get 6 moles of OH⁻. That means that you needed 6 moles of HCl since 1 mole of HCl contains 1 mole of H⁺ and equal amounts H⁺ and OH⁻ reacted with each other. To find the molarity of the HCl solution you need to divide 6mol by 1L to get 6M. Tat means that the concentration of the acid was 6M.
I hope this helps. Let me know if anything was unclear.
The completed and balanced equation is:
<span>Hg(N<span>O3</span><span>)2</span>+N<span>a2</span>Cr<span>O4</span>=HgCr<span>O4</span>+2Na(N<span>O3</span>)
Hope this helps :)</span>
<u>Answer:</u> The weight fraction of ethanol in the mixture is
<u>Explanation:</u>
We are given:
Mass of water = 1 kg = 2.205 lb (Conversion factor: 1 kg = 2.205 lb)
Mass of ethanol = 1.9 lb
Mass of ethyl acetate = 4.6 lb
Mass of mixture = [2.205 + 1.9 + 4.6] = 8.705 lb
To calculate the percentage composition of ethanol in mixture, we use the equation:

Mass of mixture = 8.705 lb
Mass of ethanol = 1.9 lb
Putting values in above equation, we get:

Hence, the weight fraction of ethanol in the mixture is 21.8 %